PHYS 1441 – Section 501
Lecture #5
Wednesday, June 16, 2004
Dr. Jaehoon Yu
•
Chapter four: Newton’s Laws of Motion
–
–
Newton’s Third Law of Motion
Solving problems using Newton’s Laws
•
•
Free-body diagram
Uniform circular motion
Today’s homework is homework #3, due 5pm, next Friday!!
Wednesday, June 16, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
1
Newton’s Third Law (Law of Action and Reaction)
If two objects interact, the force, F21, exerted on object 1 by object 2
is equal in magnitude and opposite in direction to the force, F12,
exerted on object 1 by object 2.
F21
F12
F 12   F 21
The action force is equal in magnitude to the reaction force but in
opposite direction. These two forces always act on different objects.
What is the reaction force to the
force of a free fall object?
The force exerted by the ground
when it completed the motion.
Stationary objects on top of a table has a reaction force (normal force)
from table to balance the action force, the gravitational force.
Wednesday, June 16, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
Example of Newton’s 3rd Law
A large man and a small boy stand facing each other on frictionless ice. They put their
hands together and push against each other so that they move apart. a) Who moves away
with the higher speed and by how much?
F12
M
F21=-F12
m
b) Who moves farther while
their hands are in contact?
F 12  F 21  F
F 12   F 21
ur
r
F 12 ma b F 12 x mabx F 12 y  maby  0
ur
r
F 21  M a M F 21x  MaMx F 21 y  MaMy  0
ur
ur ur
F M
ur
a
bx 
 aMx
F

12
 F 21  F
F 12   F 21
m m
vMxf  vMxi  aMxt  aMx t
vbxf  vbxi  abxt  abxt 
 vbxf  vMxf if M  m by the ratio of the masses
Given in the same time interval, since the boy
has higher acceleration and thereby higher
speed, he moves farther than the man.
Wednesday, June 16, 2004
M
M
aMxt 
vMxf
m
m
xb  vbxf t 
xb 
M
m
1
M
M
abxt 2 
vMxf t 
aMxt 2
2
m
2m
M
1

2 
 vMxf t  aMx t   m xM
2


PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
3
Some Basic Information
When Newton’s laws are applied, external forces are only of interest!!
Why?
Because, as described in Newton’s first law, an object will keep its
current motion unless non-zero net external force is applied.
Normal Force, n:
Reaction force that reacts to gravitational
force due to the surface structure of an object.
Its direction is perpendicular to the surface.
Tension, T:
The reactionary force by a stringy object
against an external force exerted on it.
Free-body diagram
Wednesday, June 16, 2004
A graphical tool which is a diagram of external
forces on an object and is extremely useful analyzing
forces and motion!! Drawn only on an object.
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
4
Free Body Diagrams and Solving Problems
•

1.
2.
3.
4.
5.
6.

Free-body diagram: A diagram of vector forces acting on an object
A great tool to solve a problem using forces or using dynamics
Select a point on an object and w/ information given
Identify all the forces acting only on the selected object
Define a reference frame with positive and negative axes specified
Draw arrows to represent the force vectors on the selected point
Write down net force vector equation
Write down the forces in components to solve the problems
No matter which one we choose to draw the diagram on, the results should be the same,
as long as they are from the same motion
FN
M
FN
Which one would you like to select to draw FBD?
What do you think are the forces acting on this object?
Gravitational force
FG  M g
Gravitational force
Me
m
The force pulling the elevator (Tension)
What about the box in the elevator?
Wednesday, June 16, 2004
FG  M g
A force supporting the object exerted by the floor
Which one would you like to select to draw FBD?
What do you think are the forces acting on this elevator?
FT
FN
F GB  mg
FG  M g
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
Gravitational
force
Normal
force
FT
FG  M g
FN
F5BG  m g
Applications of Newton’s Laws
Suppose you are pulling a box on frictionless ice, using a rope.
M
What are the forces being
exerted on the box?
T
Gravitational force: Fg
n= -Fg
Free-body
diagram
n= -Fg
T
Fg=Mg
Normal force: n
T
Fg=Mg
Tension force: T
Total force:
F=Fg+n+T=T
If T is a constant
force, ax, is constant
Wednesday, June 16, 2004
 Fx  T  Ma x
F
y
ax 
  Fg  n  Ma y  0
T
M
ay  0
v xf  vxi  axt  v xi  
T 
t
M 
1 T  2
x

x

v
t


t
x  f i xi
2M 
PHYS 1441-004,What
Springhappened
2004
to the motion in y-direction?
Dr. Jaehoon Yu
6
Example of Using Newton’s Laws
A traffic light weighing 125 N hangs from a cable tied to two other cables
fastened to a support. The upper cables make angles of 37.0o and 53.0o with
the horizontal. Find the tension in the three cables.
37o
y
53o
T1
Free-body
Diagram
T2
37o
53o
T3
r
ur ur ur ur
F  T 1  T 2  T 3  ma  0
i 3
x-comp. of
Tix  0
net force Fx  
i 1
y-comp. of
net force
i 3
Fy   Tiy  0
i 1
Wednesday, June 16, 2004
x
Newton’s 2nd law
 
  T  cos 37  T
T sin  37   T sin  53   mg  0
T sin 53   0.754  sin 37   1.25T  125N
cos 53o
 T1 cos 37  T2 cos 53  0


o
1
1
o
2
0.754T2
o
2


2
2
T2  100 N ; T1  0.754T2  75.4 N
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
7
Example w/o Friction
A crate of mass M is placed on a frictionless inclined plane of angle q.
a) Determine the acceleration of the crate after it is released.
y
n
x
q
F  Fg  n  ma
n
Fx  Ma x  Fgx  Mg sin q
Free-body
Diagram
q
Fg
y
Supposed the crate was released at the
top of the incline, and the length of the
incline is d. How long does it take for the
crate to reach the bottom and what is its
speed at the bottom?
ax  g sin q
x
F= -Mg F  Ma  n  F  n  mg cos q  0
y
y
gy
1 2 1
v
t

a x t  g sin q t 2  t 
d  ix
2
2
v xf  vix  axt g sin q
2d
g sin q
2d
 2dg sin q
g sin q
 vxf  2dg sin q
Wednesday, June 16, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
8
Forces of Friction
Resistive force exerted on a moving object due to viscosity or other types
frictional property of the medium in or surface on which the object moves.
These forces are either proportional to velocity or normal force
Force of static friction, fs: The resistive force exerted on the object until
just before the beginning of its movement
Empirical
Formula
f s  s n
What does this
formula tell you?
Frictional force increases till it
reaches to the limit!!
Beyond the limit, there is no more static frictional force but kinetic
frictional force takes it over.
Force of kinetic friction, fk
fk  k n
Wednesday, June 16, 2004
The resistive force exerted on the object
during its movement
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
9
Example w/ Friction
Suppose a block is placed on a rough surface inclined relative to the horizontal. The
inclination angle is increased till the block starts to move. Show that by measuring
this critical angle, qc, one can determine coefficient of static friction, s.
y
n
Free-body
Diagram
Fg
n
fs=kn
x
F= -Mg
q
q
Net force
F  Ma  Fg n f s
x comp.
Fx  Fgx  f s  Mg sin q  f s  0
f s   s n  Mg sin q c
y comp.
Fy  Ma y  n  Fgy  n  Mg cosqc  0
n Fgy  Mg cosq c
Mg sin q c
Mg sin q c
 tan q c
s 

Mg cos q c
n
Wednesday, June 16, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
10
Newton’s Second Law & Uniform Circular Motion
m
Fr
The centripetal acceleration is always perpendicular
to velocity vector, v, for uniform circular motion.
v2
ar 
r
r
Fr
Are there forces in this motion? If so, what do they do?
The force that causes the centripetal acceleration
acts toward the center of the circular path and
causes a change in the direction of the velocity
vector. This force is called centripetal force.
v2
 Fr  mar m r
What do you think will happen to the ball if the string that holds the ball breaks? Why?
Based on Newton’s 1st law, since the external force no longer exist, the ball will
continue its motion without change and will fly away following the tangential
direction to the circle.
Wednesday, June 16, 2004
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
11
Example of Uniform Circular Motion
A ball of mass 0.500kg is attached to the end of a 1.50m long cord. The ball is
moving in a horizontal circle. If the string can withstand maximum tension of 50.0 N,
what is the maximum speed the ball can attain before the cord breaks?
Fr m
Centripetal
acceleration:
When does the
string break?
v2
ar 
r
v2
 Fr  mar  m r  T
when the centripetal force is greater than the sustainable tension.
v2
m
 T
r
v  Tr  50.0 1.5  12.2  m / s 
m
0.500
Calculate the tension of the cord
when speed of the ball is 5.00m/s.
Wednesday, June 16, 2004
v2
 5.00   8.33 N
 0.500 
T m
 
r
1.5
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
2
12
Example of Banked Highway
(a) For a car traveling with speed v around a curve of radius r, determine a formula
for the angle at which a road should be banked so that no friction is required to keep
the car from skidding.
y
x
2
mv
x comp.  Fx  n sin q  mar  n sin q 
 0
2
r
mv
n sin q 
r
n cos q  mg
y comp.  Fy  n cos q  mg  0
mg
n
sin q
mg sin q
mv 2
n sin q 
 mg tan q 
cos q
r
v2
tan q 
gr
(b) What is this angle for an expressway off-ramp curve of radius 50m at a design
speed of 50km/h?
v  50km / hr  14m / s
Wednesday, June 16, 2004
tan q 
142
50  9.8
 0.4
PHYS 1441-004, Spring 2004
Dr. Jaehoon Yu
q  tan 1 0.4  22o
13